Question

\(\left\{{}\begin{matrix}x+y=3000\\1.1x+1.12y=3328\end{matrix}\right.\)

Answer 1

\(\left\{{}\begin{matrix}x+y=3000\\1,1x+1,12y=3328\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}1,1x+1,1y=3300\\1,1x+1,12y=3328\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=3000\\-0,02y=-28\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1400=3000\\y=\dfrac{-28}{-0,02}=1400\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3000-1400\\y=1400\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1600\\y=1400\end{matrix}\right.\)

Vậy: \(x=1600,y=1400\)